Question: Divide the following complex numbers. $\dfrac{8+19i}{4-3i}$
Solution: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate, which is ${4+3i}$. $ \dfrac{8+19i}{4-3i} = \dfrac{8+19i}{4-3i} \cdot \dfrac{{4+3i}}{{4+3i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$. $ = \dfrac{(8+19i) \cdot (4+3i)} {4^2 - (-3i)^2} $ Evaluate the squares in the denominator and subtract them. $ = \dfrac{(8+19i) \cdot (4+3i)} {(4)^2 - (-3i)^2} $ $ = \dfrac{(8+19i) \cdot (4+3i)} {16 + 9} $ $ = \dfrac{(8+19i) \cdot (4+3i)} {25} $ The denominator now doesn't contain any imaginary unit multiples, so it is a real number. Note that when a complex number, $a + bi$ is multiplied by its conjugate, the product is always $a^2 + b^2$. Now, we can multiply out the two factors in the numerator. $ \dfrac{({8+19i}) \cdot ({4+3i})} {25} $ $ = \dfrac{{8} \cdot {4} + {19} \cdot {4 i} + {8} \cdot {3 i} + {19} \cdot {3 i^2}} {25} $ $ = \dfrac{32 + 76i + 24i + 57 i^2} {25} $ Finally, simplify the fraction. $ \dfrac{32 + 76i + 24i - 57} {25} = \dfrac{-25 + 100i} {25} = -1+4i $